A pendulum is swinging next to a wall. The distance from the bob of the pendulum to the wall varies in a periodic way that can be modeled by a trigonometric function. The modeling function has period $0.8$ seconds, amplitude $6 \text{ cm}$, and midline $H = 15 \text{ cm}$. At time $t = 0.2$, the bob is at its midline, moving towards the wall. Find the formula of the trigonometric function that models the distance $H$ between the bob and the wall after $t$ seconds. Define the function using radians. $ H(t) = $ What is the distance from the pendulum to the wall after $0.5$ seconds? Round your answer, if necessary, to two decimal places. $ $
Solution: Let's start by writing a formula for the distance from the bob to the wall $u$ seconds after it passes its midline $(u$ seconds after the time $t = 0.2)$. Both sine and cosine can be used to model periodic contexts. We can decide which is better fitting by considering the $y$ -intercept. The sine function intercepts the $y$ -axis at its midline, and the cosine function intercepts the $y$ -axis at its peak. Since the bob is at its midline at time $u = 0$, we can use a sine function to model its distance, since sine functions also pass their midline at $u = 0$. The bob's distance from the wall is decreasing at time $u= 0$, while $\sin u$ is increasing at time $u= 0$, so we'll have to flip $\sin u$ vertically. Since the ordinary sine function $f(u) = \sin u$ has period $2\pi$, midline $y = 0$, and amplitude $1$, we can stretch it horizontally by a factor of ${\dfrac{0.8}{2\pi}}$, stretch it vertically by a factor of ${6}$ and flip it vertically, and move it up ${15}$ units: $ H(u) = {-6}\sin\left({\dfrac{2\pi}{0.8}}u\right) + {15}$ Since the bob passes its midline $0.2$ seconds after the stopwatch is started, $t$ seconds after the stopwatch is started is $t - 0.2$ seconds after it passes its midline, so $u = t - 0.2$ : $ H(t) = {-6}\sin\left({\dfrac{2\pi}{0.8}}(t-0.2)\right) + {15}$ When $t = 0.5$, the distance from the bob to the wall is $ \begin{aligned}H(0.5) &= {-6}\sin\left({\dfrac{2\pi}{0.8}}(0.5-0.2)\right) + {15}\\&\approx 10.76\end{aligned}$ A correct formula for $H(t)$ is: $ H(t) = -6\sin\left(\dfrac{2\pi}{0.8}(t-0.2)\right) + 15$ The distance between the bob and the wall after $0.5$ seconds is: $ 10.76\text{ cm}$ This is the graph of the function, with the bob's distance after $0.5$ seconds marked.